Smooth a positive function on a smooth manifold Let $f:M\to\Bbb R$ be a positive function on a smooth manifold $M$ such that there is no convergent sequence of points $x_k\in M$ such that $f(x_k)\to 0$. Is there a smooth (or at least continuous) function $g:M\to\Bbb R$ such that $0<g(x)<f(x)$ for any $x\in M$?
 A: First claim, that you seem to know: for all compact $K\subset M$, there exists $c>0$ such that $f(x) > c$ on $K$, for the opposite would imply the existence of a sequence $(x_n)$ in $K$ such that $f(x_n)<1/n$, and extracting a convergent subsequence would contradict your hypothesis.
Second claim, which is a basic property of differentiable manifold: there exists a locally finite covering of open sets $(U_\alpha)_{\alpha\in I}$ such that the closure of $U_\alpha$ is compact. Indeed differentiable manifolds are metrizable and so paracompact, which means that every open cover has an locally finite refinement. Then, choosing for each $x\in M$ a neighborhood $U_x$ such that the closure of $U_x$ is compact, a locally finite refinement of the covering $(U_x)_{x\in M}$ satisfies the claim.
Finally, you can choose a smooth partition of unity $(\chi_\alpha)_{\alpha\in I}$ subordinate to this open cover. According to the first claim, there exists a family of positive real numbers $(c_\alpha)_{\alpha\in I}$ such that on $U_\alpha$, $f(x)>c_\alpha>0$, and you can choose $g(x)=\sum_{\alpha} c_\alpha\chi_\alpha(x)$.
